Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r^2 - 81}{-10r - 80} \div \dfrac{-2r - 18}{r + 8} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{r^2 - 81}{-10r - 80} \times \dfrac{r + 8}{-2r - 18} $ First factor the quadratic. $q = \dfrac{(r + 9)(r - 9)}{-10r - 80} \times \dfrac{r + 8}{-2r - 18} $ Then factor out any other terms. $q = \dfrac{(r + 9)(r - 9)}{-10(r + 8)} \times \dfrac{r + 8}{-2(r + 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (r + 9)(r - 9) \times (r + 8) } { -10(r + 8) \times -2(r + 9) } $ $q = \dfrac{ (r + 9)(r - 9)(r + 8)}{ 20(r + 8)(r + 9)} $ Notice that $(r + 8)$ and $(r + 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ \cancel{(r + 9)}(r - 9)(r + 8)}{ 20(r + 8)\cancel{(r + 9)}} $ We are dividing by $r + 9$ , so $r + 9 \neq 0$ Therefore, $r \neq -9$ $q = \dfrac{ \cancel{(r + 9)}(r - 9)\cancel{(r + 8)}}{ 20\cancel{(r + 8)}\cancel{(r + 9)}} $ We are dividing by $r + 8$ , so $r + 8 \neq 0$ Therefore, $r \neq -8$ $q = \dfrac{r - 9}{20} ; \space r \neq -9 ; \space r \neq -8 $